In the univariate case, for fixed H ∈(0,1), the law of FBM is unique up to a constant. This follows in a standard way by usingH-self-similarity and stationarity of the increments
as EX(s)X(t) = 1 2(EX(t) 2+EX(s)2−E(X(t)−X(s))2) = EX(1)2 2 {|t| 2H+|s|2H − |t−s|2H}.
The same arguments cannot be applied in the case of OFBM. In fact, for OFBM,
EX(t)X(s)∗+EX(s)X(t)∗
=EX(t)X(t)∗+EX(s)X(s)∗−E(X(t)−X(s))(X(t)−X(s))∗ =|t|HΓ(1,1)|t|H∗+|s|HΓ(1,1)|s|H∗− |t−s|HΓ(1,1)|t−s|H∗,
and it is not true in general thatEX(t)X(s)∗=EX(s)X(t)∗. In this sense, a given operator
H does not characterize the law of OFBM. This also does not exclude the case where two different Hs lead to the same OFBM, and we will see in Section 4.5 below that this may happen.
Even though a fixedH does not determine the law of OFBM, an alternative character- ization can be sought through integral representations of OFBMs. In the univariate case, it is well-known that FBM has the spectral representation
BH(t) = 1 C2(H) Z R eixt−1 ix |x| −(H−1/2)Be(dx), (4.6)
where Be(x) =Be1(x) +iBe2(x) is a complex-valued Brownian motion such thatBe1(−x) =
e
B1(x) and Be2(−x) = −Be2(x), and C2(H) is a normalizing constant (see, for instance,
Samorodnitsky and Taqqu (1994), p. 328). The representation (4.6) also yields the law of FBM, and sheds light on its structure (that is, it says how it can be built from the usual BM). It is therefore natural to try to obtain integral representations for OFBMs. This is done through a number of results given next.
Definition 4.3.1. We will say a function f : R → Cn is operator-homogeneous of degree
K ∈M(n) if, forc >0,
As with ordinary homogeneity, all operator-homogeneous functions of the same degree differ only by an operator constant, which is their value on the sphere. In fact, from (4.7),
f(c) =cKf(1) =:cKA, c >0, A∈M(n,C),
f(c) = (−c)Kf(−1) =: (−c)KB, c <0, B∈M(n,C).
For our purposes, the value off at zero is defined arbitrarily. Consequently, any operator- homogeneous function of degree K can be written in the form
f(x) =xK+A+xK−B. (4.8)
In Theorem 4.3.1 below, we establish integral representations of OFBMs in the spectral domain. Before that, we state a technical lemma.
Lemma 4.3.1. Let {Ye(x)}x∈R∈Cn be an orthogonal-increment process, and setF
ij(dx) =
EYei(dx)Yej(dx), wherei, j= 1, ..., nand Yei is a component ofYe. IfFii(dx) andFjj(dx) are
absolutely continuous with respect to the Lebesgue measure over a given interval, then so is
Fij(dx).
Proof. A consequence of the Cauchy-Schwartz Inequality.
Theorem 4.3.1. Let {BH(t)}t∈R be OFBM with o.s.s. exponentH, where the real parts of
the characteristic roots of H are in the interval (0,1). Then,
{BH(t)}t∈R=d n Z R eitx−1 ix (x −D + A+x−−DA)dBe(x) o t∈R, (4.9)
where D = H−I(1/2), A ∈ GL(n,C), A is the matrix whose entries are the complex conjugates of the entries ofA, andBe(x) :=Be1(x) +iBe2(x)is a complex-valued multivariate Brownian motion satisfyingBe1(−x) =Be1(x), Be2(−x) =−Be2(x)and EdBe(x)dBe(x)∗ =dx. Proof. For notational simplicity, set X=BH. Since X has stationary increments, we have
X(t)−X(s) =
Z
eitx−eisx
where Ye(dx) is an orthogonal-increment random measure in Cn (see Doob (1990)). Since
Re(hk) > 0 for all k = 1, ..., n, then X(0) = 0 a.s. (see Maejima and Mason (1994)).
Therefore, X can be represented as
X(t) =
Z
R
eitx−1
ix Ye(dx). (4.11)
Moreover, since X is Gaussian,Ye(dx) is a Gaussian random measure. Let
FX(dx) =EYe(dx)Ye(dx)∗
be the multivariate spectral distribution ofYe(dx). The rest of the proof goes in three steps: (i) showing the existence of a spectral density function,
(ii) decorrelating the measure Ye(dx) by finding a filter based upon the spectral density function,
(iii) showing that the filter is an operator-homogeneous function.
Step (i): SinceX is o.s.s. with exponentH,
X(ct)=d cH
Z
R
eitx−1
ix Ye(dx). (4.12)
On the other hand, through a change of variables v=cx,
X(ct)=d Z R eitv−1 iv cYe(c −1dv). (4.13)
In differential form, this means that
cHYe(dx)=d cYe(c−1dx) (4.14)
or, equivalently,
e
Thus,FX([0, c]) can be written as
EYe([0, c])Ye([0, c])∗=cI−HFX([0,1])(cI−H)∗, (4.16)
forc >0 without loss of generality. By Lemma 4.3.1, it suffices to prove that the individual
Fii are absolutely continuous. By the explicit formula for cI−H, the individual entries
FX([0, c])ij in the expression on the right-hand side of (4.16) are linear combinations (with
complex weights) of terms of the form
(log(c))l
l! c
1−hk, k= 1, ..., n, l∈N (4.17)
(or their respective conjugate), or identically zero for c > 0. Thus, FX(c) is differentiable inc over (0,∞) since FX([0, c])ij =FX(c)ij−FX(0)ij.
We want to prove that Ye(0) = 0 a.s. We now proceed as in Maejima and Mason (1994). Since the real part of the eigenvalues of I−H are strictly greater than zero, then by Proposition 2.1.(ii) in Maejima and Mason (1994) we havektI−H k →0 ast→0, where
k.kis the (complex) operator norm. Thus, by equation (4.15),
kYe(0)k =d kcI−HYe(0)k ≤ kcI−H k kYe(0)k →0 as t→0.
So,Ye(0) = 0 a.s., as claimed. Now note that
e Y(c)=d cI−HY(1), (4.18) and since kcI−HYe(1)k≤kcI−H k kYe(1)k →0 as c→0, we also have cI−HYe(1)→0 as c→0 (4.19)
imply that
e
Y(c)→L2 0 =Ye(0) as c→0 (i.e.,Ye isL2-stochastically continuous at zero). Therefore,
FX([−c, c])→0 as c→0, because FX([−c, c]) =E ³ Z c −c dYe(x) ´³ Z c −c dYe(x) ´∗ =E ³ Z 0 −c dYe(x) + Z c 0 dYe(x) ´³ Z 0 −c dYe(x) + Z c 0 dYe(x) ´∗ =E ³ Z 0 −c dYe(x) ´³ Z 0 −c dYe(x) ´∗ +E ³ Z c 0 dYe(x) ´³ Z c 0 dYe(x) ´∗ →0 as c→0,
where the third equality follows by the orthogonal increments ofYe. This implies that
FX(0)−FX(0−) = lim
c→0FX(c)−FX(−c) = limc→0FX([−c, c]) = 0.
As a consequence, for all i= 1, ..., n we have thatFii(c) is differentiable forc6= 0 and
continuous at zero. Thus, a multivariate spectral density functionfX(x) exists.
Step (ii): Since fX(x) is positive definite Hermitian-symmetric for everyx, the Spectral Theorem yields a square root ba(x) of fX(x). Let dBe(x) be a complex-valued multivariate Brownian motion as in the statement of the theorem. The random measure ba(x)dBe(x) is equal (in distribution) toYe(dx), since
E(ba(x)dBe(x)dBe(x)∗ba(x)∗) =ba(x)ba(x)∗dx=fX(x)dx=FX(dx).
This implies thatX can also be represented as
X(t)=d
Z
R
eitx−1
Step (iii): By rewriting (4.14) with ba(x)dBe(x), we obtain that cHba(x)dBe(x)=d ba ³x c ´ cI(1/2)dBe(x), whence b a(cx) =c−Dba(x). (4.21)
This means that ba is operator-homogeneous of degree K = −D, which implies it has the form (4.8) and representation (4.9) holds.
We next obtain integral representations of OFBMs in the time domain. We will use the following elementary result. We write f ∈ L2(R,Rn2
) for a matrix-valued function f
when RR[f(u)◦f(u)]du < +∞, where A◦B := trace(A∗B). The Fourier transform of
f ∈L2(R,Rn2
) is defined as fb(x) =RRe−ixuf(u)du.
Lemma 4.3.2. Let f, g∈L2(R,Rn2
). Then, the Plancherel identity holds, i.e.,
Z R f(u)g(u)∗du= 1 2π Z R b f(x)bg(x)∗dx, (4.22)
where fband bg are the component-wise Fourier transforms off and g. Theorem 4.3.2. Let {BH(t)}t∈R be OFBM with o.s.s. exponent H. Then,
{BH(t)}t∈R=d n Z R ³ ((t−u)D+−(−u)D+)M+ ((t−u)D−−(−u)D−)N ´ dB(u) o t∈R, (4.23) where D =H−I(1/2), (M, N) ∈ (GL(n)∪ {0})×(GL(n)∪ {0})\{(0,0)} and B(u) is a real-valued, multivariate Brownian motion.
Proof. Let X and Xe denote the processes on the right-hand side of (4.9) and (4.23), re- spectively. It suffices to show that the covariance structures ofX and Xe are the same. For simplicity, we only considerXe in (4.23) withN = 0 and show that it has the representation (4.9).
As in the univariate case, one can show that Z R ³ (t−u)D+−(−u)D+ ´
e−iuxdu= (e−itx−1)|x|−(D+I)Γ(D+I)eisign(x)π(D+I)/2,
where
Γ(K) =
Z ∞
0
e−xxK−Idx
converges absolutely if the characteristic roots of the operator K are greater than zero. Then, by Lemma 4.3.2, EXe(s)Xe(t)∗ = Z R ((s−u)D+−(−u)D+)M M∗((t−u)D+∗−(−u)D+∗)du = 1 2π Z R (e−isx−1)(eitx−1) |x|2 (|x| −DΓ(D+I)eisign(x)π(D+I)/2)M M∗· ·(e−isign(x)π(D∗+I)/2Γ(D+I)∗|x|−D∗)dx.
This is also the covariance structure forX in (4.9) with A:= Γ(D+I)eiπ(D+I)/2M. Note also that for Xe to take values in Rn, it is necessary that (M, N) ∈ (GL(n)∪ {0})×(GL(n)∪ {0})\{(0,0)}. In fact, any operators M ,f Ne ∈ M(n,C) have the form
f
M = M1 +iM2, Ne = N1+iN2, where M1, M2, N1, N2 ∈ M(n) (actually, we must have
M1 or N1 ∈ GL(n), otherwise Xe cannot be a proper process in Rn). By considering the
expression (4.23) withM :=Mf,N :=Ne, it follows thatXe(t)∈Rnfor a giventif and only
ifRR
³
((t−u)D
+−(−u)D+)M2+ ((t−u)D−−(−u)D−)N2
´
dB(u) = 0, which does not hold a.s. unlessM2 =N2 = 0.
Remark 4.3.1. For what operatorsH is the time domain representation (4.23) of OFBM well-defined? Let D = H −(1/2)I. The integral (4.23) is well-defined as long as the integrand is in L2(R). Using the Jordan form of D=P JP−1, where P ∈GL(n,C) and J
is in Jordan normal form with the eigenvalues dl,l= 1, ..., nof D, the square-integrability
follows if|t−u|J− | −u|J is inL2(R). By Appendix B.2, it is enough to have the functions
in L2(R), where m = 1, ..., n
Jdl and nJdl is the size of the Jordan block Jdl of D. The functions (4.24) are inL2(R) whendl∈(−1/2,1/2).